$\frac{x}{x+1}=1-\frac{1}{x+1}; \frac{y}{y+1}=1-\frac{1}{y+1}; \frac{z}{z+4}=1-\frac{4}{z+4}\Rightarrow VT=3-(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+1}) \leftrightarrow \frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\geq \frac{(1+1+2)^{2}}{x+y+z+1+1+4}\geq \frac{8}{3}\Rightarrow VT\leq 3-\frac{8}{3}= \frac{1}{3}$
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